All categories

3 years ago

Donna buys 2 pies for 6 people to share. How much of a pie does each person get? Find the solution.

Mathematics

1 answer:

Fittoniya [83]3 years ago

4 0

Answer:

1/3 each

Step-by-step explanation:

2= 2/1 6= 6/1

2/1 divided into 6/1 is the same as 2/1 multiplied by 1/6

2/1 x 1 /6 = 2/6 = 1/3

You might be interested in

A number line going from 1 to 13. 1 dot is above 1. 3 dots are above 2. 1 dot is above 3. 3 dots are above 4. 1 dot is above 5.

aliina [53]

Answer:

Answer: C) The mean will increase more than the median, but both will increase.

Step-by-step explanation:

I got it right on the test review.

5 0

3 years ago

Read 2 more answers

What is 4.2 kilometers in meters

vovikov84 [41]

THE ANSWER IS 42,000.

TO GET THIS WE KNOW THAT 1 KILOMETERS IS 10,000.

SO YOU TAKE 4.2 AND MULTIPLY IT BY 10,000. THAT EQUALS 42,000.

4 0

3 years ago

Please help they are separate questions please help me with this!

monitta

Answer: Mary makes $1160, Steven makes $142.50

Step-by-step explanation:

Facts

MARY

8% commision on everything she sells

She sells a car for $14,500

The first question about Mary asks how much commision she makes. To find that, lets find 8% of 14,500

8% of 14,500= 0.08x14,500= 1160

Therefore, Mary makes $1,160 in commision.

STEVEN

Facts

He makes 15% in commision for the supplies he sells

He sells $950 in supplies

Money made =?

15% of 950=0.15x950=$142.50

6 0

2 years ago

Use series to verify that<br><br> <img src="https://tex.z-dn.net/?f=y%3De%5E%7Bx%7D" id="TexFormula1" title="y=e^{x}" alt="y=e^{

SVETLANKA909090 [29]

$y = e^x\\\\\displaystyle y = \sum_{k=1}^{\infty}\frac{x^k}{k!}\\\\\displaystyle y= 1+x+\frac{x^2}{2!} + \frac{x^3}{3!}+\ldots\\\\\displaystyle y' = \frac{d}{dx}\left( 1+x+\frac{x^2}{2!} + \frac{x^3}{3!}+\frac{x^4}{4!}+\ldots\right)\\\\$

$\displaystyle y' = \frac{d}{dx}\left(1\right)+\frac{d}{dx}\left(x\right)+\frac{d}{dx}\left(\frac{x^2}{2!}\right) + \frac{d}{dx}\left(\frac{x^3}{3!}\right) + \frac{d}{dx}\left(\frac{x^4}{4!}\right)+\ldots\\\\\displaystyle y' = 0+1+\frac{2x^1}{2*1} + \frac{3x^2}{3*2!} + \frac{4x^3}{4*3!}+\ldots\\\\\displaystyle y' = 1 + x + \frac{x^2}{2!}+ \frac{x^3}{3!}+\ldots\\\\\displaystyle y' = \sum_{k=1}^{\infty}\frac{x^k}{k!}\\\\\displaystyle y' = e^{x}\\\\$

This shows that $y' = y$ is true when $y = e^x$

-----------------------

Note 1: A more general solution is $y = Ce^x$ for some constant C.
Note 2: It might be tempting to say the general solution is $y = e^x+C$ , but that is not the case because $y = e^x+C \to y' = e^x+0 = e^x$ and we can see that $y' = y$ would only be true for C = 0, so that is why $y = e^x+C$ does not work.